Construct a quadrilateral ABCD in which AB = 5.6 cm,

Solution:

Steps of construction:

Step 1: Draw $A B=5.6 \mathrm{~cm}$

Step 2: Make $\angle A=50^{\circ}$ and $\angle B=105^{\circ}$

Step 3: With $B$ as the centre, draw an arc of $4 \mathrm{~cm}$.

Step 3: Sum of all the angles of the quadrilateral is $360^{\circ}$.

$\angle A+\angle B+\angle C+\angle D=360^{\circ}$

$50^{\circ}+105^{\circ}+\angle C+80^{\circ}=360^{\circ}$

$235^{\circ}+\angle C=360^{\circ}$

$\angle C=360^{\circ}-235^{\circ}$

$\angle C=125^{\circ}$

Step 5: With $C$ as the centre, make $\angle C$ equal to $\angle 125^{\circ}$.

Step 6: Join $C$ and $D$.

Step 7: Measure $\angle D=80^{\circ}$

Then, $A B C D$ is the required quadrilateral.